Step 1

To find how many matricec A of order 2 exists such that \(A^2=I\)

Here we have to find \(A^2=A \times A =I\)

Let us consider any arbitrary matrix of order 2. \(A=\begin{pmatrix}a & b \\c & d \end{pmatrix}\)

Then \(A^2\) is given by \(A^2=\begin{pmatrix}a & b \\c & d \end{pmatrix} \begin{pmatrix}a & b \\c & d \end{pmatrix} = \begin{pmatrix}a^2+bc & ab+bd \\ca+dc & d^2+bc \end{pmatrix}\)

Now from \(A^2=I\) , we get \(A^2 = \begin{pmatrix}a^2+bc & b(a+d) \\c(a+d) & d^2+bc \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} =I\)

\((1) a^2+bc=1\)

\((2) d^2+bc=1\)

\((3) b(a+d)=0\)

\((3) c(a+d)=0\)

Step 2

Case 1:

If \((a+d) \neq 0\)

From (3) we get : \(b=0\)

From (4) we get : \(c=0\)

From (1) we get : \(a^2=1 \Rightarrow a=\pm1\)

From (2) we get : \(d^2=1 \Rightarrow d=\pm1\)

Therefore we get the matrix of the form

\(\left\{A=\begin{pmatrix}a & 0 \\0 & d \end{pmatrix} : a=\pm1 , d= \pm1 \right\}\) Therefore there are only 4 matrices exist such that \(A^2=I\) from case 1.

Step 3

Case 2:

If \((a+d)=0\)

sub case a:

If a=0 , that implies \(d=0\)

Then from(1 ) and (2) , we get : \(bc=1 \Rightarrow c=b^{-1}=\frac{1}{b}\) in real numbers.

Therefore we get the matrix of the form

\(\left\{A=\begin{pmatrix} 0 & a \\a^{-1} & 0 \end{pmatrix} :a \in F \right\}\) Therefore there are infinitely many matrices exist if the field is the set of real numbers such that \(A^2=I\)

Step 4

We can also consider further subcases to find the type of involuntary matrices, but as the question is asked to find how many, we have already got infinitely many \(2 \times 2\) matrices of such type.

Similarly for any \(n>2\) , we can find infinitely many matrices satisfying this condition on the field of real numbers.